# Sum of binomial coefficients and cos(kx)

• flyerpower
In summary, @vela suggested using Euler's formula to solve for the sum of a series, but @rglv said a simpler approach is to use a multiple angle formula for cos(nx). @vela found the answer using euler's formula.
flyerpower

## Homework Statement

Calculate the following sum:
(click to expand)

## The Attempt at a Solution

I tried something with Moivre formula and Newton binomial theorem but no result , should i continue with these or is there any simpler approach?

I just need some hints.

Thanks.

cos(k*x) = (1/2)[exp(i*k*x) + exp(-i*k*x)], where i = sqrt(-1), and exp(i*k*x) = z^k for an appropriate z.

RGV

Ray Vickson said:
cos(k*x) = (1/2)[exp(i*k*x) + exp(-i*k*x)], where i = sqrt(-1), and exp(i*k*x) = z^k for an appropriate z.

RGV

Thanks for the hint but i cannot still work it out.

I plugged cos(kx) directly in the sum and resulted in some exponentials..

[URL]http://www2.wolframalpha.com/Calculate/MSP/MSP256719g79b7bde68e6h6000020i46ai95ba26aii?MSPStoreType=image/gif&s=20&w=225&h=36[/URL]

The result should look like this : [URL]http://www2.wolframalpha.com/Calculate/MSP/MSP126419g79d95ifaec8fe00001ifd1gdagh48hf7e?MSPStoreType=image/gif&s=3&w=152&h=36[/URL]

May i get one more hint ?:p

Last edited by a moderator:
Try a multiple angle formula for cos(nx) such as:
$$\cos(n x) = \cos\left({{n x}\over{3}}\right) \left(-1+2 \cos\left({{2 (n x)}\over{3}}\right)\right)$$
or
$$\cos(n x) = -1+2\cos^2\left({{n x}\over{2}}\right)$$

It's easier to use the fact that cos nx = real part of einx. The idea is to calculate a related sum where the real part is the answer you're looking for.

Ok, thank you both for the ideas, i'll try to solve it tomorrow and come back with a result.

I wasn't able to come up with a result.

Here is my trial:

i got this result using euler's formula
(e^(ix)+1)^n + (e^(-ix)+1)^n = 2 SUM(k=0,n) C{k,n} * cos(k*x)

I tried how @vela said but i couldn't figure out what sum to calculate.

flyerpower said:
I wasn't able to come up with a result.

Here is my trial:

i got this result using euler's formula
(e^(ix)+1)^n + (e^(-ix)+1)^n = 2 SUM(k=0,n) C{k,n} * cos(k*x)

I tried how @vela said but i couldn't figure out what sum to calculate.

I don't understand your comment. You have already computed the sum, so you already have the answer. However, maybe the person who posed the problem does not want to see complex quantities, and in that case you need to extract the real part of (1+exp(ix))^n, which you can do by converting 1 + exp(ix) to polar form, for example.

RGV

flyerpower said:
I wasn't able to come up with a result.

Here is my trial:

i got this result using euler's formula
(e^(ix)+1)^n + (e^(-ix)+1)^n = 2 SUM(k=0,n) C{k,n} * cos(k*x)
Use the fact that
\begin{eqnarray*}
e^{ix}+1 &= e^{ix/2}(e^{ix/2}+e^{-ix/2}) \\
e^{-ix}+1 &= e^{-ix/2}(e^{ix/2}+e^{-ix/2})
\end{eqnarray*}
I tried how @vela said but i couldn't figure out what sum to calculate.
Using my suggestion would have given you
$$\sum_{k=0}^n \begin{pmatrix}n \\ k\end{pmatrix} \cos kx = \sum_{k=0}^n \begin{pmatrix}n \\ k\end{pmatrix} \mathrm{Re}(e^{ikx}) = \mathrm{Re}\left[\sum_{k=0}^n \begin{pmatrix}n \\ k\end{pmatrix} e^{ikx}\right] = \cdots$$

Well this is too much :), I'm lost in the imaginary world, i'll give it a try another time.
Your help is appreciated, thank you.

You'll find Euler's formula
$$e^{i\theta} = \cos\theta + i\sin\theta$$
very helpful in general. Using it, you can express the trig functions in terms of complex exponentials
\begin{align*}
\cos\theta &= \frac{e^{i\theta}+e^{-i\theta}}{2} \\
\sin\theta &= \frac{e^{i\theta}-e^{-i\theta}}{2i}
\end{align*}
Back to your problem... You don't have much left to do to get the answer. For example, you have
$$(e^{ix}+1)^n = [e^{ix/2}(e^{ix/2}+e^{-ix/2})]^n = e^{inx/2}(e^{ix/2}+e^{-ix/2})^n$$
Nothing terribly complicated happening there, just algebra. Now if you compare what's inside the parentheses to the formulas above, you can see or show it's equal to
$$e^{inx/2}(e^{ix/2}+e^{-ix/2})^n = e^{inx/2}[2\cos(x/2)]^n$$
Now simplify the term $(e^{-ix}+1)^n$ similarly and then combine the results.

Thank you for the explanation!

## 1. What is the sum of binomial coefficients?

The sum of binomial coefficients, also known as the binomial theorem, is a mathematical formula that allows you to expand the power of a binomial expression. It is expressed as (a + b)^n = ∑[k = 0 to n] (n choose k) * a^(n-k) * b^k, where (n choose k) represents the binomial coefficient.

## 2. How is the sum of binomial coefficients related to Pascal's triangle?

Pascal's triangle is a triangular array of numbers that are used to calculate binomial coefficients. Each row of the triangle represents the coefficients of the binomial expansion for a specific power. The sums of the coefficients in each row correspond to the powers of 2, making it a useful tool for calculating binomial coefficients.

## 3. Can the sum of binomial coefficients be simplified?

Yes, the sum of binomial coefficients can be simplified using various mathematical techniques such as the binomial theorem or the Pascal's triangle. These methods allow you to find patterns and simplifications in the coefficients, making it easier to calculate the sum.

## 4. What is the significance of cos(kx) in the sum of binomial coefficients?

The cosine function, cos(kx), is used in the sum of binomial coefficients to represent the alternating signs of the coefficients. This is because the binomial coefficients follow a pattern of alternating positive and negative signs, and the cosine function allows us to express this pattern mathematically.

## 5. How is the sum of binomial coefficients used in real-world applications?

The sum of binomial coefficients has various applications in fields such as statistics, probability, and physics. It is used to calculate the probabilities of different outcomes in experiments and to model physical phenomena such as particle interactions. It is also used in engineering and computer science for tasks such as error correction and data compression.

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